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NCERT Solutions · Class 12 · Chapter 10: Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.3

Exercise 10.3: Scalar (dot) product. \(\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3\). Perpendicular exactly when the dot product is 0. Projection of \(\vec a\) on \(\vec b\): \(\dfrac{\vec a \cdot \vec b}{|\vec b|}\). \(\vec a \cdot \vec a = |\vec a|^2\), so expand \((\vec a + \vec b) \cdot (\vec a + \vec b)\) like an algebraic square.

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Exercise 10.3 questions and solutions

Exercise 10.3, Question 1

Find the angle between \(\vec a\) and \(\vec b\) with \(|\vec a| = \sqrt3\), \(|\vec b| = 2\), \(\vec a \cdot \vec b = \sqrt6\).
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  1. \[\begin{aligned}\cos\theta &= \dfrac{\sqrt6}{2\sqrt3} \\ &= \dfrac{1}{\sqrt2}\end{aligned}\]
Answer: \(\dfrac{\pi}{4}\)

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Exercise 10.3, Question 2

Find the angle between \(\hat i - 2\hat j + 3\hat k\) and \(3\hat i - 2\hat j + \hat k\).
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  1. Dot product \(3 + 4 + 3 = 10\); both magnitudes \(\sqrt{14}\).
Answer: \(\cos^{-1}\dfrac57\)

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Exercise 10.3, Question 4

Find the projection of \(\hat i + 3\hat j + 7\hat k\) on \(7\hat i - \hat j + 8\hat k\).
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  1. \(\dfrac{7 - 3 + 56}{\sqrt{49 + 1 + 64}}\).
Answer: \(\dfrac{60}{\sqrt{114}}\)

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Exercise 10.3, Question 5

Show \(\tfrac17(2\hat i + 3\hat j + 6\hat k)\), \(\tfrac17(3\hat i - 6\hat j + 2\hat k)\), \(\tfrac17(6\hat i + 2\hat j - 3\hat k)\) are unit vectors and mutually perpendicular.
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  1. Each has components with squares summing to \(\tfrac{49}{49} = 1\).
  2. Dot products: \(6 - 18 + 12 = 0\), \(18 - 12 - 6 = 0\), \(12 + 6 - 18 = 0\) (each over 49).
Answer: Shown

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Exercise 10.3, Question 6

Find \(|\vec a|\) and \(|\vec b|\) if \((\vec a + \vec b) \cdot (\vec a - \vec b) = 8\) and \(|\vec a| = 8|\vec b|\).
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  1. \[\begin{aligned}&|\vec a|^2 - |\vec b|^2 = 8 \\ \Rightarrow\ &63|\vec b|^2 = 8\end{aligned}\]
Answer: \(|\vec b| = \dfrac{2\sqrt2}{3\sqrt7}\), \(|\vec a| = \dfrac{16\sqrt2}{3\sqrt7}\)

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Exercise 10.3, Question 7

Evaluate \((3\vec a - 5\vec b) \cdot (2\vec a + 7\vec b)\).
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  1. Expand: \[6\vec a \cdot \vec a + 21\vec a \cdot \vec b - 10\vec b \cdot \vec a - 35\vec b \cdot \vec b\]
Answer: \[6|\vec a|^2 + 11\vec a \cdot \vec b - 35|\vec b|^2\]

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Exercise 10.3, Question 8

\(\vec a\), \(\vec b\) have the same magnitude, the angle between them is \(60^\circ\) and \(\vec a \cdot \vec b = \tfrac12\). Find the magnitudes.
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  1. \[\begin{aligned}&|\vec a|^2\cos 60^\circ = \tfrac12 \\ \Rightarrow\ &|\vec a|^2 = 1\end{aligned}\]
Answer: \(|\vec a| = |\vec b| = 1\)

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Exercise 10.3, Question 9

Find \(|\vec x|\) if \(\vec a\) is a unit vector and \((\vec x - \vec a) \cdot (\vec x + \vec a) = 12\).
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  1. \[\begin{aligned}&|\vec x|^2 - |\vec a|^2 = 12 \\ \Rightarrow\ &|\vec x|^2 = 13\end{aligned}\]
Answer: \(\sqrt{13}\)

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Exercise 10.3, Question 10

\(\vec a = 2\hat i + 2\hat j + 3\hat k\), \(\vec b = -\hat i + 2\hat j + \hat k\), \(\vec c = 3\hat i + \hat j\), and \(\vec a + \lambda\vec b\) is perpendicular to \(\vec c\). Find \(\lambda\).
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  1. \[\vec a + \lambda\vec b = (2 - \lambda)\hat i + (2 + 2\lambda)\hat j + (3 + \lambda)\hat k\]
  2. Dot with \(\vec c\): \[\begin{aligned}3(2 - \lambda) + (2 + 2\lambda) &= 8 - \lambda \\ &= 0\end{aligned}\]
Answer: \(\lambda = 8\)

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Exercise 10.3, Question 11

Show \(|\vec a|\vec b + |\vec b|\vec a\) is perpendicular to \(|\vec a|\vec b - |\vec b|\vec a\) for non-zero \(\vec a, \vec b\).
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  1. Expand: \[(|\vec a|\vec b + |\vec b|\vec a) \cdot (|\vec a|\vec b - |\vec b|\vec a) = |\vec a|^2\,\vec b \cdot \vec b - |\vec a||\vec b|\,\vec b \cdot \vec a + |\vec b||\vec a|\,\vec a \cdot \vec b - |\vec b|^2\,\vec a \cdot \vec a\]
  2. The middle terms cancel (\[\vec a \cdot \vec b = \vec b \cdot \vec a\]), and \[|\vec a|^2|\vec b|^2 - |\vec b|^2|\vec a|^2 = 0\]; a zero dot product of non-zero vectors means they are perpendicular.
Answer: Shown

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Exercise 10.3, Question 12

If \(\vec a \cdot \vec a = 0\) and \(\vec a \cdot \vec b = 0\), what can be concluded about \(\vec b\)?
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  1. \(\vec a \cdot \vec a = |\vec a|^2 = 0\), so \(\vec a = \vec 0\); then \(\vec a \cdot \vec b = 0\) for every \(\vec b\).
Answer: Nothing: \(\vec b\) can be any vector

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Exercise 10.3, Question 13

\(\vec a, \vec b, \vec c\) are unit vectors with \(\vec a + \vec b + \vec c = \vec 0\). Find \(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a\).
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  1. \[\begin{aligned}|\vec a + \vec b + \vec c|^2 &= 3 + 2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a) \\ &= 0\end{aligned}\]
Answer: \(-\tfrac32\)

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Exercise 10.3, Question 14

If \(\vec a = \vec 0\) or \(\vec b = \vec 0\), then \(\vec a \cdot \vec b = 0\). Show the converse need not be true.
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  1. \(\vec a = \hat i\), \(\vec b = \hat j\): both non-zero, but \(\vec a \cdot \vec b = 0\).
Answer: e.g. \(\hat i \cdot \hat j = 0\) with both vectors non-zero

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Exercise 10.3, Question 15

A\((1, 2, 3)\), B\((-1, 0, 0)\), C\((0, 1, 2)\). Find \(\angle ABC\) (the angle between \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\)).
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  1. \[\overrightarrow{BA} = 2\hat i + 2\hat j + 3\hat k\], \[\overrightarrow{BC} = \hat i + \hat j + 2\hat k\]; dot product \(2 + 2 + 6 = 10\).
  2. \(\cos B = \dfrac{10}{\sqrt{17}\sqrt6}\).
Answer: \[\angle ABC = \cos^{-1}\dfrac{10}{\sqrt{102}}\]

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Exercise 10.3, Question 16

Show A\((1, 2, 7)\), B\((2, 6, 3)\), C\((3, 10, -1)\) are collinear.
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  1. \[\overrightarrow{AB} = \hat i + 4\hat j - 4\hat k\] and \[\overrightarrow{BC} = \hat i + 4\hat j - 4\hat k\]: parallel with a common point B.
Answer: Shown

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Exercise 10.3, Question 17

Show \(2\hat i - \hat j + \hat k\), \(\hat i - 3\hat j - 5\hat k\), \(3\hat i - 4\hat j - 4\hat k\) form the vertices of a right-angled triangle.
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  1. \[\overrightarrow{AB} = -\hat i - 2\hat j - 6\hat k\], \[\overrightarrow{BC} = 2\hat i - \hat j + \hat k\], \[\overrightarrow{CA} = -\hat i + 3\hat j + 5\hat k\]
  2. \[\begin{aligned}\overrightarrow{BC} \cdot \overrightarrow{CA} &= -2 - 3 + 5 \\ &= 0\end{aligned}\]: right angle at C (also \(6 + 35 = 41\)).
Answer: Shown (right angle at C)

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Exercise 10.3, Question 18

\(\vec a\) is non-zero with magnitude a, \(\lambda\) a non-zero scalar. \(\lambda\vec a\) is a unit vector if: (A) \(\lambda = 1\) (B) \(\lambda = -1\) (C) \(a = |\lambda|\) (D) \(a = \dfrac{1}{|\lambda|}\)
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  1. \(|\lambda\vec a| = |\lambda|a = 1\).
Answer: (D)

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Done the NCERT exercises? The board paper asks more

Vector Algebra has 39 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

Vector Algebra in our sample papers: Sample paper 1 (questions 13, 15, 24) · Sample paper 2 (questions 16, 19, 34) · Sample paper 3 (questions 13, 14, 24) · Sample paper 4 (questions 16, 19, 25) · Sample paper 5 (questions 13, 19, 24).

Also useful: free MCQs and case studies for Vector Algebra · formulas for this chapter (Class 12 formula sheet, free PDF) · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.